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horrorfan [7]
3 years ago
7

Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4 [fast] (4) C6H5SO3− + H3O+ → C6H5SO3H + H2O
Chemistry
1 answer:
ziro4ka [17]3 years ago
6 0

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

                        rate = k_{2} [SO_{3}][C_{6}H_{6}]

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

                       rate = k_{1}[H_{2}SO_{4}]^{2}

Hence when we substitute equation 2 in equation one,

                   Rate comes out to be =  k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

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Vesna [10]
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8 0
3 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}
3 0
3 years ago
How many moles of MgSiO3 are in 237g of the compound?
Anna35 [415]
For this, we first calculate molecular weight of MgSiO₃:
Atomic masses:
Mg = 24
Si = 28
O = 16

Mr = 24 + 28 + 16 x 3
Mr = 100

moles = mass / Mr
moles = 237 / 100
moles = 2.37
6 0
3 years ago
A jet airplane has a velocity of 1145 knots. A knot is 1 nautical mile (nm)/hr. A nautical
ra1l [238]

Answer:

589.038 m/s

Explanation:

i dont know if did this right tho

4 0
2 years ago
Calculate the concentration of formate in a 100mm solution of formic acid at ph 4.15. The pka for formic acid is 3.75
MissTica

The molarity of formic acid is 100 mM or 100\times 10^{-3}M. The dissociation reaction of formic acid is as follows:

HCOOH\leftrightharpoons HCOO^{-}+H^{+}

The expression for dissociation constant of the reaction will be:

K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}

Rearranging,

[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}

Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:

[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M

Similarly, pK_{a}=3.75 thus,

[K_{a}=10^{-pK_{a}}=10^{-3.75}=1.78\times 10^{-4}M

Putting the values,

[HCOO^{-}]=\frac{(1.78\times 10^{-4}M)(100\times 10^{-3}M)}{(7.08\times 10^{-5}M}=0.2511 M

Therefore, the concentration of formate will be 0.2511 M.

4 0
3 years ago
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